Question – 1

1. A 350 kg car moving with a speed of 25 m/s collides with a truck and stops. If the whole process takes 3 s, what is the force experienced by the car due to the truck?

- A. 2516.76 N
- B. 2550.67 N
- C. 2916.67 N
- D. 3016.76 N
- E. 8750 N
- Answer:C
- Answer Explanation:According to Newton’s second law of motion we have F = p/tMomentum is given as p = mvp = 350 X 25 Or p = 8750 kgm/sTherefore F= 8750/3 Or F= 2916.67 NAs the car finally stops, force experienced by the car is retarding in nature. Hence, C is the correct answer option.

Question – 2

2. If a ball is projected up in the air at an angleof45o from the horizontal with a velocity of 10 m/s, for how long will the ball be in the air?

- A. 0.5 s
- B. 1.12 s
- C. 1.44 s
- D. 2.44 s
- E. 1.12 s
- Answer:C
- Answer Explanation:The time for which the ball will be in the air is given by t = 2u sin ??/gt= 2 X 10 Xsin 45
^{o}/ 9.8 Or t = 20/ 9.8 X ???2t = 1.44sTherefore, C is the correct answer option.

Question – 3

3. If the speeds of sound in solid, liquid and gas are arranged in increasing order, which of the following is a correct arrangement?

- A. Solids < liquids < gas
- B. Liquids < solids < gas
- C. Gas < liquids < solids
- D. Gas < solids < liquids
- E. Liquids < gas < solids
- Answer:C
- Answer Explanation:The speed of sound in solids like steel and iron is 5000 m/s and 5130 m/s. The speed of sound in liquids like water at 0
^{o}C and 20^{o}C is 1402 m/s and 1482 m/s respectively. The speed of sound in air (gas)at 0^{o}C and 20^{o}C is 331.5 m/s and 344 m/s respectively. Thus, from the above discussion it can be concluded that the speed of sound in solids is more than that in liquids and air. Therefore, the correct arrangement for speeds of sound is air (gas) < liquids < solids. Thus, C is the correct answer option.

Question – 4

4. What is the current density of electrons flowing through a steel block of length 1 m, if potential difference applied across the conductor is 200 V? (Conductivity of steel = 1.45 X 10^{6} ??^{-1} m^{-1})

- A. 2.9 X 10
^{8}A/m^{2} - B. 1.4 X 10
^{8}A/m^{2} - C. 2.2 X 10
^{8}A/m^{2} - D. 3.4 X 10
^{9}A/m^{2} - E. 1.45 X 10
^{4}A/m^{2} - Answer:A
- Answer Explanation:Current density is given as J = ??EElectric field is given as E = V/ l Or E = 200/1 =200 V/mJ = 1.45 X 10
^{6}X 200 Or J = 2.9 X 10^{8}A/m^{2}Hence, A is the correct answer option.

Question – 5

5. A heat engine performs work of 10000 J while transferring heat of 12000 J. What is the efficiency of the engine?

- A. 83.3 %
- B. 84.5 %
- C. 78.6 %
- D. 45.7 %
- E. 41.6 %
- Answer:A
- Answer Explanation:Efficiency of a heat engine is given as ?? = W/Q?? = 10000/12000 Or ?? = 0.833%?? = 0.833 X 100 %%?? = 83.3%Therefore, A is the correct answer option.

Question – 6

6. Force acting on an object is given a F = Xx^{-1} (Where x is the distance by which the body is displaced due to the applied force) What would be the dimensional formula for X?

- A. MLT
^{-2} - B. MLT
^{-1} - C. M
^{-1}LT^{-1} - D. M
^{-1}L^{2}T^{-1} - E. ML
^{2}T^{-2} - Answer:E
- Answer Explanation:From dimensional analysis we know that dimensional formula of the left hand side of an expression should be the same as the dimensional formula of the individual quantities on the right hand side. Force has the SI unit of kgms
^{-2}. Therefore, the dimensional formula of force becomes MLT^{-2}. Dimensional formula of x is L. Hence, dimensional formula of x^{-1}becomes L^{-1}. As said earlier, Xx^{-1}should have the same dimensional formula as F. Therefore, it can be written as MLT^{-2}= X x L^{-1}Or X = ML^{2}T^{-2}. Therefore, E is the correct answer option.

Question – 7

7. A 5 kg block of ice, kept at some height, has the potential energy of 1000 Joules. What is the height at which the block has been kept?

- A. 20.4 m
- B. 21.2 m
- C. 10.5 m
- D. 11.4 m
- E. 24.3 m
- Answer:A
- Answer Explanation:Potential energy is given as PE = mghPutting the values in equation 1 we get1000 = 5 X 9.8 X hh = 20.4 mHence, A is the correct answer option.

Question – 8

8. A beta particle of mass 9.1 X 10^{-31} kg is moving perpendicular to a magnetic field in a region of 0.01 T with a speed of 10^{4} m/s. What is the force experienced by the beta particle due to the magnetic field?(Charge on electron = 1.6 X 10^{-19} C)

- A. 3.2 X 10
^{-17}N - B. 1.6 X 10
^{-17}N - C. 4.8 X 10
^{-17}N - D. 2.4 X 10
^{-17}N - E. 10
^{-17}N - Answer:B
- Answer Explanation:Force experienced by the beta particle is given as F = qvB sin ??F = 1.6 X 10
^{-19}X 10^{4}X 0.01 X sin 90^{o}F = 1.6 X 10^{-17}X 1 NF = 1.6 X 10^{-17}NHence, B is the correct answer option.

Question – 9

9. If a block of wood weighs 1000 N, what would be its mass?

- A. 102.04 kg
- B. 10.24 kg
- C. 1020.44 kg
- D. 1.24 kg
- E. 110.24 kg
- Answer:A
- Answer Explanation:W=mgPutting the value of W and g in equation 1 we get1000 = m X 9.8m = 1000/9.8 Or m = 102.04 kgTherefore, A is the correct answer option.

Question – 10

10. A small water drop on a surface is viewed through a glass. If the real depth of the water drop is 15 cm, what will be the apparent depth of the water drop? (Refractive index of glass =1.5 and refractive index of air = 1)

- A. 5 cm
- B. 7.5 cm
- C. 8.4 cm
- D. 1.5 cm
- E. 10 cm
- Answer:E
- Answer Explanation:The apparent depth is given as
_{glass}??^{air}= ??_{glass}/ ??_{air}= real depth / appearent depth1.5 / 1 = 15cm / apparent depthapparent depth = 15 cm / 1.5 Or apparent depth = 10 cmHence, E is the correct answer option.

Question – 11

11. What would be the specific gravity of a fluid having density of 944 kg/m^{3}?

- A. 0.967
- B. 0.456
- C. 0.944
- D. 0.789
- E. 0.45
- Answer:C
- Answer Explanation:SG = ??
_{final}/ ??_{water}SG = 944 / 1000SG = 0.944Therefore, C is the correct answer option.

Question – 12

12. If a wheel is rotating with an angular velocity of , what is the time taken by the wheel to complete one full rotation?

- A. 1.57 s
- B. 2.51 s
- C. 1.84 s
- D. 3.52 s
- E. 2.22 s
- Answer:B
- Answer Explanation:?? = 2?? / T2.5 = 2 X 3.14 / TT = 2 X 3.14 / 2.5 Or T = 2.51 sTherefore, B is the correct answer option.

Question – 13

13. What is the Joule equivalent of 1 eV?(Charge on electron = 1.6 X 10^{-19} C)

- A. 1.6 X 10
^{-19}J - B. 3.2 X 10
^{-19}J - C. 3.2 X 10
^{-13}J - D. 1.6 X 10
^{-13}J - E. 1 J
- Answer:A
- Answer Explanation:Joule is the SI unit of energy. eV is also a unit of energy but for microscopic charged particles. One electron volt(eV) is defined as the energy possessed by an electron moving in a unit potential field. We know that energy of a charged particle in a potential field is given asE = qV JFor an electron in a potential field of 1 V, energy can be expressed by using equation asE = eV J1 eV = 1.6 X 10
^{-19}X 1 J Or 1 eV = 1.6 X 10^{-19}JTherefore, A is the correct answer option.

Question – 14

14. A block of mass M, connected to a spring of spring constant k (a spring-mass system),is executing SHM with time period, T on Earth. If the same system executes SHM on the surface of Moon, what will be the new time period?

- A. More than that on the Earth
- B. Less than that on the Earth
- C. Same as that on the Earth
- D. Zero
- E. Not enough data is provided to predict the answer
- Answer:C
- Answer Explanation:Time period is given as T = 2?? &radic(M/k)At the surface of the Moon, acceleration due to gravity is 1/6 th of that on the surface of the Earth.As, there is no term related to acceleration due to gravity in equation 1, time period of the system will remain unaffected. Hence, time period of the system on the moon will be the same as that of the system on the Earth. Therefore, C is the correct answer option.

Question – 15

15. What is the dipole moment between an electron and a proton separated by a distance of 1 nm?

- A. 3.6 X 10
^{-28}Cm - B. 4.5 X 10
^{-28}Cm - C. 4.5 X 10
^{-26}Cm - D. 1.6 X 10
^{-28}Cm - E. 3.2 X 10
^{-28}Cm - Answer:D
- Answer Explanation:Electric dipole moment between two charges of opposite polarity is given as P = QdThe distance is in nm so we need to convert it into m.1 nm = 10
^{-9}mP = 1.6 X 10^{-19}X 10^{-9}Or P = 1.6 X 10^{-28}CmHence, D is the correct answer option.